Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}x+9y &= 8 \\ 4x+6y &= 2\end{align*}$
Solution: Begin by moving the $y$ -term in the second equation to the right side of the equation. $4x = -6y+2$ Divide both sides by $4$ to isolate $x$ $x = {-\dfrac{3}{2}y + \dfrac{1}{2}}$ Substitute this expression for $x$ in the first equation. $({-\dfrac{3}{2}y + \dfrac{1}{2}}) + 9y = 8$ $-\dfrac{3}{2}y + \dfrac{1}{2} + 9y = 8$ Simplify by combining terms, then solve for $y$ $\dfrac{15}{2}y + \dfrac{1}{2} = 8$ $\dfrac{15}{2}y = \dfrac{15}{2}$ $y = 1$ Substitute $1$ for $y$ in the top equation. $x+9( 1) = 8$ $x+9 = 8$ $x = -1$ The solution is $\enspace x = -1, \enspace y = 1$.